NOI模拟（5.14） JSOID1T3 绝地反击（bzoj5316）

绝地反击

题目背景：

5.14 模拟 JSOI2018D1T3

分析：计算几何 + 网络流 + 二分 + 网络流退流

又是我不会的新姿势，首先我们考虑如何暴力，对于最优的时间，我们可以肯定，一定有一个飞船是恰好在这个时间到了对应坐标，而没有任何停留，否则一定可以更优，那么我们二分最大时间，然后枚举每一条飞船作为最晚到达的那一个，然后就可以确定对应的位置，然后跑一遍二分图匹配即可，这样做的话普通写大概有50 ~ 60，加上一些特判和一些剪枝大概能够有90左右，当然，这个结果是loj得到的。

时间复杂度：O(n^7/2logn~ n⁴logn)，据说用HK可以卡过去？

Source:

/*created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <bitset>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;
}///*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = read(), iosig = false; !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true; }for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
//*/const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN];
char *oh = obuf;
inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;
}template<class T>
inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}
}inline void flush() {for (int i = 0; i <= 10; ++i) std::cout << obuf[i];std::cout << '\n';fwrite(obuf, 1, oh - obuf, stdout), oh = obuf;
}/*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = getchar(), iosig = false; !isdigit(c); c = getchar())if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
//*/const int MAXN = 400 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int INF = MAXN;struct node {int to, w, rev;node() {}node(int to, int w, int rev) : to(to), w(w), rev(rev) {}
} ;std::vector<node> edge[MAXN << 1];
int dis[MAXN], temp[MAXN];
inline void add_edge(int x, int y, int w) {edge[x].push_back(node(y, w, edge[y].size()));edge[y].push_back(node(x, 0, edge[x].size() - 1));
}inline bool bfs(int s, int t) {static std::queue<int> q;memset(dis, -1, sizeof(int) * (t + 5));memset(temp, 0, sizeof(int) * (t + 5));dis[s] = 0, q.push(s);while (!q.empty()) {int cur = q.front();q.pop();for (int p = 0; p < edge[cur].size(); ++p) {node *e = &edge[cur][p];if (e->w && dis[e->to] == -1) dis[e->to] = dis[cur] + 1, q.push(e->to);}}return dis[t] != -1;
}inline int dfs(int cur, int low, int t) {if (cur == t) return low;int delta = 0;for (int &p = temp[cur]; p < edge[cur].size(); ++p) {node *e = &edge[cur][p];if (e->w && dis[e->to] == dis[cur] + 1) {int ret = dfs(e->to, std::min(low - delta, e->w), t);e->w -= ret, edge[e->to][e->rev].w += ret;if ((delta += ret) == low) break ;}}return delta;
}inline int dinic(int s, int t) {int ans = 0, ret = 0;while (bfs(s, t)) do ret = dfs(s, INF, t); while (ans += ret, ret);return ans;
}struct point {double x, y;point() {}point(double x, double y) : x(x), y(y) {}inline point operator + (const point &a) const {return point(x + a.x, y + a.y);}inline point operator - (const point &a) const {return point(x - a.x, y - a.y);}inline point operator / (const double a) const {return point(x / a, y / a);}inline point operator * (const double a) const {return point(x * a, y * a);}inline double dis() {return sqrt(x * x + y * y);}inline double dis2() {return x * x + y * y;}inline point rotate(double ang) {double cos_a = cos(ang), sin_a = sin(ang);return point(x * cos_a - y * sin_a, x * sin_a + y * cos_a);}
} p[MAXN], st[MAXN];int n, r, x, y;inline point get_pos(point p, double dis, double flag) {double a = dis, b = r, c = p.dis();double cos_a = (b * b + c * c - a * a) / 2.00 / b / c;point cur = p / c * r;return cur = cur.rotate(flag * acos(cos_a)), cur;
}inline bool check(point st, int id, double max) {int s = 0, t = 2 * n + 1;for (int i = s; i <= t; ++i) edge[i].clear();for (int i = 1; i < n; ++i) {st = st.rotate(2.0 * PI / n);int cc = 0;for (int j = 1; j <= n; ++j) if (j != id && (p[j] - st).dis2() <= max * max) add_edge(j, i + n, 1), cc++;if (cc == 0) return false;}for (int i = 1; i <= n; ++i) if (i != id) add_edge(s, i, 1);for (int i = 1; i < n; ++i) add_edge(i + n, t, 1);return (dinic(s, t) == n - 1);
}inline bool check(double x) {for (int i = 1; i <= n; ++i) {if (x < p[i].dis() - r || x < r - p[i].dis()) return false;if (x > p[i].dis() + r) {st[i] = point(INF, INF);continue ;}st[i] = get_pos(p[i], x, 1);}for (int i = 1; i <= n; ++i)if (st[i].x != INF && check(st[i], i, x)) return true;return false;
}inline void solve() {R(n), R(r);for (int i = 1; i <= n; ++i) R(x), R(y), p[i] = point(x, y);std::random_shuffle(p + 1, p + n + 1);double l = -eps, r = 0;for (int i = 1; i <= n; ++i) r = std::max(p[i].dis() + ::r, r);int cnt = 0;while (l + eps < r) {double mid = (l + r) / 2;check(mid) ? r = mid : l = mid;}printf("%0.8f", r);
}int main() {// freopen("in.in", "r", stdin);// freopen("fleet.out", "w", stdout);solve();return 0;
}

考虑如何优化复杂度，我们发现如果把x轴正半轴上的(r, 0)当做其中一个坐标，那么所有的可行方案与这个方案的角度都不超过2pi / n，那么我们考虑维护逆时针进行偏转的过程，我们发现，如果一开始先将(r, 0)定位为1号点，那么随着偏转，每一个点原有的可行匹配点会在某一时刻减少一个，也会在某一时刻增加一个，那么我们可以将每一种操作的偏转角度记录下来，然后将操作排序，显然操作总个数是O(n)的，每一次操作相当于从网络流建图中删掉一条边，或者增加一条边，我们只需要将这个流量退掉，或者加上就可以了，因为是二分图，所以一次只会影响三条边，还是很轻松的，因为流量改变只有一，所以一次退流或者加边的复杂度都应该是O(n + m)的，那么总复杂度都是O(n³logn)。

Source:

/*created by scarlyw
*/
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cctype>
#include <vector>
#include <set>
#include <queue>
#include <ctime>
#include <deque>
#include <iterator>
#include <map>inline char read() {static const int IN_LEN = 1024 * 1024;static char buf[IN_LEN], *s, *t;if (s == t) {t = (s = buf) + fread(buf, 1, IN_LEN, stdin);if (s == t) return -1;}return *s++;
}// /*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = read(), iosig = false; !isdigit(c); c = read()) {if (c == -1) return ;if (c == '-') iosig = true;   }for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
//*/const int OUT_LEN = 1024 * 1024;
char obuf[OUT_LEN], *oh = obuf;
inline void write_char(char c) {if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf;*oh++ = c;
}template<class T>
inline void W(T x) {static int buf[30], cnt;if (x == 0) write_char('0');else {if (x < 0) write_char('-'), x = -x;for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;while (cnt) write_char(buf[cnt--]);}
}inline void flush() {fwrite(obuf, 1, oh - obuf, stdout);
}/*
template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) {if (c == -1) return ;if (c == '-') iosig = true; }for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}
// */const int MAXN = 200000 + 10;
const double PI = acos(-1.0);
const double eps = 1e-9;
const int INF = MAXN;double block;
int s, t, n, r, x, y;
int dis[MAXN], temp[MAXN];
bool vis[MAXN];struct point {double x, y;point() {}point(double x, double y) : x(x), y(y) {}inline double dis() {return sqrt(x * x + y * y);}inline double dis2() {return x * x + y * y;}
} p[MAXN];struct node {int to, w, rev;node() {}node(int to, int w, int rev) : to(to), w(w), rev(rev) {}
} ;std::vector<node> edge[MAXN];struct data {double num;int u, v, type;data() {}data(double num, int u, int v, int type): num(num), u(u), v(v), type(type) {}inline bool operator < (const data &a) const {return (num == a.num) ? (type > a.type) : (num < a.num);}
} q[MAXN << 1 | 1];inline void add_edge(int x, int y, int w) {edge[x].push_back(node(y, w, edge[y].size()));edge[y].push_back(node(x, 0, edge[x].size() - 1));
}inline bool bfs(int s, int t) {static std::queue<int> q;for (int i = s; i <= t; ++i) dis[i] = -1;for (int i = s; i <= t; ++i) temp[i] = 0;dis[s] = 0, q.push(s);while (!q.empty()) {int cur = q.front();q.pop();for (int p = 0; p < edge[cur].size(); ++p) {node *e = &edge[cur][p];if (e->w > 0 && dis[e->to] == -1) dis[e->to] = dis[cur] + 1, q.push(e->to);}}return dis[t] != -1;
}inline int dfs(int cur, int low, int t) {if (cur == t) return low;int delta = 0;for (int &p = temp[cur]; p < edge[cur].size(); ++p) {node *e = &edge[cur][p];if (dis[e->to] == dis[cur] + 1 && e->w > 0) {int ret = dfs(e->to, std::min(low - delta, e->w), t);e->w -= ret, edge[e->to][e->rev].w += ret;if ((delta += ret) == low) break ;}}return delta;
}int ret, ans;inline void pop_flow(int u, int v) {std::vector<node>::iterator it, it1;bool flag;for (it = edge[u].begin(); it != edge[u].end(); ++it) if (it->to == v) {flag = it->w, it = edge[u].erase(it);break ;} for (; it != edge[u].end(); ++it) edge[it->to][it->rev].rev--;for (it = edge[v].begin(); it != edge[v].end(); ++it)if (it->to == u) {it = edge[v].erase(it);break ;}for (; it != edge[v].end(); ++it) edge[it->to][it->rev].rev--;if (flag) return ;ans--;for (it = edge[s].begin(); it != edge[s].end(); ++it)if (it->to == u) {it->w ^= 1, edge[it->to][it->rev].w ^= 1;break ;}for (it = edge[t].begin(); it != edge[t].end(); ++it)if (it->to == v) {it->w ^= 1, edge[it->to][it->rev].w ^= 1;break ;}while (bfs(s, t))do ret = dfs(s, INF, t);while (ans += ret, ret);
}inline bool check(double x) {int cnt = 0;for (int i = s; i <= t; ++i) edge[i].clear();for (int i = 1; i <= n; ++i) {double dis = p[i].dis();if (dis > x + r || dis < r - x) return false;if (dis + r <= x) for (int j = 1; j <= n; ++j) add_edge(i, j + n, 1);else {double angle1 = atan2(p[i].y, p[i].x);double angle2 = acos(((double)r * r + dis * dis - x * x) / 2.0 / dis / r);double al = angle1 - angle2, ar = angle1 + angle2;while (al < 0) al += 2.0 * PI;while (ar < 0) ar += 2.0 * PI;int pl = al / block, pr = ar / block;q[++cnt] = data(al - block * pl, i, pl + 1 + n, 1), pl++;q[++cnt] = data(ar - block * pr, i, pr + 1 + n, 0), pr++;if (al <= ar) for (int j = pl + 1; j <= pr; ++j) add_edge(i, j + n, 1);else {for (int j = pl + 1; j <= n; ++j) add_edge(i, j + n, 1);for (int j = 1; j <= pr; ++j) add_edge(i, j + n, 1);}} }for (int i = 1; i <= n; ++i) add_edge(s, i, 1), add_edge(i + n, t, 1);std::sort(q + 1, q + cnt + 1), ret = ans = 0;while (bfs(s, t))do ret = dfs(s, INF, t);while (ans += ret, ret);if (ans == n) return true;for (int i = 1; i <= cnt; ++i) {if (q[i].type) {add_edge(q[i].u, q[i].v, 1);while (bfs(s, t))do ret = dfs(s, INF, t);while (ans += ret, ret);if (ans == n) return true;} else pop_flow(q[i].u, q[i].v);}return false;
}inline void solve() {R(n), R(r), block = PI * 2.00 / n, s = 0, t = 2 * n + 1;for (int i = 1; i <= n; ++i) R(x), R(y), p[i] = point(x, y);double l = -eps, r = 300;while (l + eps < r) {double mid = (l + r) / 2;(check(mid)) ? r = mid : l = mid;}printf("%0.8f", r);
}int main() {freopen("in.in", "r", stdin);solve();return 0;
}

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